A115057 & A000914

[Leroy Quet]

s(n+1,m) = n*s(n,m) + s(n,m-1).
s(n+1,n) = the nth triangular # = n(n+1)/2.
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Date: Fri, 12 Oct 2007 11:26:39 -0700
To: zak seidov <zakseidov at yahoo.com>, seqfan at ext.jussieu.fr
From: "T. D. Noe" <noe at sspectra.com>
Subject: Re: A115057  & A000914
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Date: Sat, 13 Oct 2007 00:40:40 +0200
From: Artur <grafix at csl.pl>
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Subject: A107751
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I would like to ask is somebody which understand A107751 and is able 
write Mathematica procedure. I was obtained that same sequence in 
completely another problem and connection is very unexpected  I want 
check that these two are that same. My procedure is on number of factors 
of  polynomials (x^n+x+1) :
Table[Length[FactorList[x^n + x + 1]] - 1, {n, 0, 200}]
{1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1,
2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 
2, 1,
1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 
1, 2,
1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 
1, 1,
2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 
2, 1,
1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 
1, 2,
1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 
1, 1,
2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2}

I will be greatfull for procedure or explanation that these two 
different matters are that same

Best wishes
Artur